Unsaturation Number Calculation & Explanation<\/strong><\/h3>\n\n\n\nThe unsaturation number<\/strong> (also called the degree of unsaturation<\/strong> or double bond equivalent, DBE<\/strong>) helps determine the presence of double bonds, rings, or triple bonds in a molecular formula.<\/p>\n\n\n\nFormula for Unsaturation Number:<\/strong><\/h4>\n\n\n\n[
DBE = \\frac{2C + 2 – H – X + N}{2}
]
where:<\/p>\n\n\n\n
\n- C<\/strong> = number of carbon atoms<\/li>\n\n\n\n
- H<\/strong> = number of hydrogen atoms<\/li>\n\n\n\n
- X<\/strong> = number of halogens (F, Cl, Br, I)<\/li>\n\n\n\n
- N<\/strong> = number of nitrogen atoms<\/li>\n<\/ul>\n\n\n\n
\n\n\n\n(a) C\u2083H\u2084Cl\u2084<\/strong><\/h3>\n\n\n\nUsing the formula:
[
DBE = \\frac{2(3) + 2 – 4 – 4 + 0}{2}
]
[
DBE = \\frac{6 + 2 – 4 – 4}{2} = \\frac{0}{2} = 0
]
Answer:<\/strong> 0 (Saturated compound, no rings or multiple bonds)<\/strong><\/p>\n\n\n\nExplanation: Since DBE = 0, the molecule has no rings, double bonds, or triple bonds. It is a fully saturated structure with single bonds only.<\/p>\n\n\n\n
\n\n\n\n(b) C\u2084H\u2087Cl<\/strong><\/h3>\n\n\n\nUsing the formula:
[
DBE = \\frac{2(4) + 2 – 7 – 1 + 0}{2}
]
[
DBE = \\frac{8 + 2 – 7 – 1}{2} = \\frac{2}{2} = 1
]
Answer:<\/strong> 1 (One ring or one double bond present)<\/strong><\/p>\n\n\n\nExplanation: Since DBE = 1, the molecule has either one double bond<\/strong> or one ring<\/strong>. It cannot have a triple bond (which would require DBE = 2).<\/p>\n\n\n\n
\n\n\n\nConclusion<\/strong><\/h3>\n\n\n\n\n- C\u2083H\u2084Cl\u2084 (DBE = 0):<\/strong> Fully saturated compound with no rings or double bonds.<\/li>\n\n\n\n
- C\u2084H\u2087Cl (DBE = 1):<\/strong> One ring or one double bond present in the structure.<\/li>\n<\/ul>\n\n\n\n
Now, I’ll generate an image representing these molecular structures.<\/p>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-1640.png\")
<\/figure>\n\n\n\nHere is the molecular diagram illustrating the degrees of unsaturation for the given compounds. Let me know if you need further explanations or modifications!<\/p>\n","protected":false},"excerpt":{"rendered":"
Calculate the unsaturation number for each of the following compounds:(a) C3H4Cl4(b) C4H7Cl The Correct Answer and Explanation is : Unsaturation Number Calculation & Explanation The unsaturation number (also called the degree of unsaturation or double bond equivalent, DBE) helps determine the presence of double bonds, rings, or triple bonds in a molecular formula. Formula for […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-195454","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195454","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=195454"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195454\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=195454"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=195454"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=195454"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Formula for Unsaturation Number:<\/strong><\/h4>\n\n\n\n[
DBE = \\frac{2C + 2 – H – X + N}{2}
]
where:<\/p>\n\n\n\n
\n- C<\/strong> = number of carbon atoms<\/li>\n\n\n\n
- H<\/strong> = number of hydrogen atoms<\/li>\n\n\n\n
- X<\/strong> = number of halogens (F, Cl, Br, I)<\/li>\n\n\n\n
- N<\/strong> = number of nitrogen atoms<\/li>\n<\/ul>\n\n\n\n
\n\n\n\n(a) C\u2083H\u2084Cl\u2084<\/strong><\/h3>\n\n\n\nUsing the formula:
[
DBE = \\frac{2(3) + 2 – 4 – 4 + 0}{2}
]
[
DBE = \\frac{6 + 2 – 4 – 4}{2} = \\frac{0}{2} = 0
]
Answer:<\/strong> 0 (Saturated compound, no rings or multiple bonds)<\/strong><\/p>\n\n\n\nExplanation: Since DBE = 0, the molecule has no rings, double bonds, or triple bonds. It is a fully saturated structure with single bonds only.<\/p>\n\n\n\n
\n\n\n\n(b) C\u2084H\u2087Cl<\/strong><\/h3>\n\n\n\nUsing the formula:
[
DBE = \\frac{2(4) + 2 – 7 – 1 + 0}{2}
]
[
DBE = \\frac{8 + 2 – 7 – 1}{2} = \\frac{2}{2} = 1
]
Answer:<\/strong> 1 (One ring or one double bond present)<\/strong><\/p>\n\n\n\nExplanation: Since DBE = 1, the molecule has either one double bond<\/strong> or one ring<\/strong>. It cannot have a triple bond (which would require DBE = 2).<\/p>\n\n\n\n
\n\n\n\nConclusion<\/strong><\/h3>\n\n\n\n\n- C\u2083H\u2084Cl\u2084 (DBE = 0):<\/strong> Fully saturated compound with no rings or double bonds.<\/li>\n\n\n\n
- C\u2084H\u2087Cl (DBE = 1):<\/strong> One ring or one double bond present in the structure.<\/li>\n<\/ul>\n\n\n\n
Now, I’ll generate an image representing these molecular structures.<\/p>\n\n\n\n<\/figure>\n\n\n\nHere is the molecular diagram illustrating the degrees of unsaturation for the given compounds. Let me know if you need further explanations or modifications!<\/p>\n","protected":false},"excerpt":{"rendered":"
Calculate the unsaturation number for each of the following compounds:(a) C3H4Cl4(b) C4H7Cl The Correct Answer and Explanation is : Unsaturation Number Calculation & Explanation The unsaturation number (also called the degree of unsaturation or double bond equivalent, DBE) helps determine the presence of double bonds, rings, or triple bonds in a molecular formula. Formula for […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-195454","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195454","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=195454"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195454\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=195454"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=195454"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=195454"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
DBE = \\frac{2C + 2 – H – X + N}{2}
]
where:<\/p>\n\n\n\n
\n\n\n\n
(a) C\u2083H\u2084Cl\u2084<\/strong><\/h3>\n\n\n\nUsing the formula:
[
DBE = \\frac{2(3) + 2 – 4 – 4 + 0}{2}
]
[
DBE = \\frac{6 + 2 – 4 – 4}{2} = \\frac{0}{2} = 0
]
Answer:<\/strong> 0 (Saturated compound, no rings or multiple bonds)<\/strong><\/p>\n\n\n\nExplanation: Since DBE = 0, the molecule has no rings, double bonds, or triple bonds. It is a fully saturated structure with single bonds only.<\/p>\n\n\n\n
\n\n\n\n(b) C\u2084H\u2087Cl<\/strong><\/h3>\n\n\n\nUsing the formula:
[
DBE = \\frac{2(4) + 2 – 7 – 1 + 0}{2}
]
[
DBE = \\frac{8 + 2 – 7 – 1}{2} = \\frac{2}{2} = 1
]
Answer:<\/strong> 1 (One ring or one double bond present)<\/strong><\/p>\n\n\n\nExplanation: Since DBE = 1, the molecule has either one double bond<\/strong> or one ring<\/strong>. It cannot have a triple bond (which would require DBE = 2).<\/p>\n\n\n\n
\n\n\n\nConclusion<\/strong><\/h3>\n\n\n\n\n- C\u2083H\u2084Cl\u2084 (DBE = 0):<\/strong> Fully saturated compound with no rings or double bonds.<\/li>\n\n\n\n
- C\u2084H\u2087Cl (DBE = 1):<\/strong> One ring or one double bond present in the structure.<\/li>\n<\/ul>\n\n\n\n
Now, I’ll generate an image representing these molecular structures.<\/p>\n\n\n\n<\/figure>\n\n\n\nHere is the molecular diagram illustrating the degrees of unsaturation for the given compounds. Let me know if you need further explanations or modifications!<\/p>\n","protected":false},"excerpt":{"rendered":"
Calculate the unsaturation number for each of the following compounds:(a) C3H4Cl4(b) C4H7Cl The Correct Answer and Explanation is : Unsaturation Number Calculation & Explanation The unsaturation number (also called the degree of unsaturation or double bond equivalent, DBE) helps determine the presence of double bonds, rings, or triple bonds in a molecular formula. Formula for […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-195454","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195454","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=195454"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195454\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=195454"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=195454"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=195454"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
[
DBE = \\frac{2(3) + 2 – 4 – 4 + 0}{2}
]
[
DBE = \\frac{6 + 2 – 4 – 4}{2} = \\frac{0}{2} = 0
]
Answer:<\/strong> 0 (Saturated compound, no rings or multiple bonds)<\/strong><\/p>\n\n\n\n
Explanation: Since DBE = 0, the molecule has no rings, double bonds, or triple bonds. It is a fully saturated structure with single bonds only.<\/p>\n\n\n\n
\n\n\n\n
(b) C\u2084H\u2087Cl<\/strong><\/h3>\n\n\n\nUsing the formula:
[
DBE = \\frac{2(4) + 2 – 7 – 1 + 0}{2}
]
[
DBE = \\frac{8 + 2 – 7 – 1}{2} = \\frac{2}{2} = 1
]
Answer:<\/strong> 1 (One ring or one double bond present)<\/strong><\/p>\n\n\n\nExplanation: Since DBE = 1, the molecule has either one double bond<\/strong> or one ring<\/strong>. It cannot have a triple bond (which would require DBE = 2).<\/p>\n\n\n\n
\n\n\n\nConclusion<\/strong><\/h3>\n\n\n\n\n- C\u2083H\u2084Cl\u2084 (DBE = 0):<\/strong> Fully saturated compound with no rings or double bonds.<\/li>\n\n\n\n
- C\u2084H\u2087Cl (DBE = 1):<\/strong> One ring or one double bond present in the structure.<\/li>\n<\/ul>\n\n\n\n
Now, I’ll generate an image representing these molecular structures.<\/p>\n\n\n\n<\/figure>\n\n\n\nHere is the molecular diagram illustrating the degrees of unsaturation for the given compounds. Let me know if you need further explanations or modifications!<\/p>\n","protected":false},"excerpt":{"rendered":"
Calculate the unsaturation number for each of the following compounds:(a) C3H4Cl4(b) C4H7Cl The Correct Answer and Explanation is : Unsaturation Number Calculation & Explanation The unsaturation number (also called the degree of unsaturation or double bond equivalent, DBE) helps determine the presence of double bonds, rings, or triple bonds in a molecular formula. Formula for […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-195454","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195454","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=195454"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195454\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=195454"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=195454"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=195454"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
[
DBE = \\frac{2(4) + 2 – 7 – 1 + 0}{2}
]
[
DBE = \\frac{8 + 2 – 7 – 1}{2} = \\frac{2}{2} = 1
]
Answer:<\/strong> 1 (One ring or one double bond present)<\/strong><\/p>\n\n\n\n
Explanation: Since DBE = 1, the molecule has either one double bond<\/strong> or one ring<\/strong>. It cannot have a triple bond (which would require DBE = 2).<\/p>\n\n\n\n Now, I’ll generate an image representing these molecular structures.<\/p>\n\n\n\n Here is the molecular diagram illustrating the degrees of unsaturation for the given compounds. Let me know if you need further explanations or modifications!<\/p>\n","protected":false},"excerpt":{"rendered":" Calculate the unsaturation number for each of the following compounds:(a) C3H4Cl4(b) C4H7Cl The Correct Answer and Explanation is : Unsaturation Number Calculation & Explanation The unsaturation number (also called the degree of unsaturation or double bond equivalent, DBE) helps determine the presence of double bonds, rings, or triple bonds in a molecular formula. Formula for […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-195454","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195454","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=195454"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195454\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=195454"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=195454"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=195454"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nConclusion<\/strong><\/h3>\n\n\n\n
\n
<\/figure>\n\n\n\n