{"id":248824,"date":"2025-07-08T18:44:33","date_gmt":"2025-07-08T18:44:33","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=248824"},"modified":"2025-07-08T18:44:36","modified_gmt":"2025-07-08T18:44:36","slug":"which-compound-matches-the-ir-spectrum-4","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/08\/which-compound-matches-the-ir-spectrum-4\/","title":{"rendered":"Which compound matches the IR spectrum"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Which compound matches the IR spectrum? 1,5-hexadiene dipropyl ether hexen-3-ol cyclohexane trans-4-octene 3-hexanol 1-methylcyclopentene<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The correct compound that matches the IR spectrum from the options provided is <strong>1,5-hexadiene<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">To identify the compound that matches an IR spectrum, we need to look for key functional groups and bond vibrations that would be indicated in the spectrum:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>1,5-hexadiene<\/strong>:\n<ul class=\"wp-block-list\">\n<li>This compound contains a conjugated system of double bonds (C=C), which typically produces distinctive absorption peaks in the IR spectrum around 1650 cm\u207b\u00b9, indicative of the C=C stretch.<\/li>\n\n\n\n<li>The diene nature (two double bonds) will also likely result in weaker absorptions around 925-980 cm\u207b\u00b9 due to out-of-plane C-H bending vibrations.<\/li>\n\n\n\n<li>The C-H stretches in the alkane region will appear around 2850-3000 cm\u207b\u00b9.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Dipropyl ether<\/strong>:\n<ul class=\"wp-block-list\">\n<li>This compound contains an ether functional group (-O-), which typically shows a strong C-O stretch around 1050-1150 cm\u207b\u00b9. It does not have conjugated double bonds, so it would not exhibit the typical peaks for alkenes or dienes.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Hexen-3-ol<\/strong>:\n<ul class=\"wp-block-list\">\n<li>This compound contains both an alcohol (-OH) and a C=C double bond. The alcohol group would show a broad peak around 3200-3550 cm\u207b\u00b9 due to the O-H stretch, and the C=C stretch would appear around 1650 cm\u207b\u00b9. However, the IR spectrum might not have the same distinct features as 1,5-hexadiene due to the presence of the alcohol group.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Cyclohexane<\/strong>:\n<ul class=\"wp-block-list\">\n<li>This is a saturated cyclic hydrocarbon, so its IR spectrum would primarily feature C-H stretches (around 2850-3000 cm\u207b\u00b9) and bending vibrations around 1450 cm\u207b\u00b9. It lacks any functional groups like double bonds or alcohols that would produce significant IR peaks.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Trans-4-octene<\/strong>:\n<ul class=\"wp-block-list\">\n<li>The trans-alkene would have a C=C stretch around 1650 cm\u207b\u00b9, but the absence of conjugation and the different structure make this less likely to match the IR spectrum of 1,5-hexadiene.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>3-hexanol<\/strong>:\n<ul class=\"wp-block-list\">\n<li>This alcohol would show a broad O-H stretch around 3200-3550 cm\u207b\u00b9 and a C-H stretch in the alkane region, but it does not have the conjugated system that would produce the unique features seen in 1,5-hexadiene.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>1-methylcyclopentene<\/strong>:\n<ul class=\"wp-block-list\">\n<li>This compound has a C=C stretch around 1650 cm\u207b\u00b9, but the structure of a cyclopentene would produce different bending vibrations compared to the open-chain diene structure of 1,5-hexadiene.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion:<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The presence of conjugated C=C double bonds in <strong>1,5-hexadiene<\/strong> is the key feature that makes it the best match for the given IR spectrum. The IR peaks associated with conjugation and specific C-H bending patterns distinguish it from the other compounds.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner8-325.jpeg\" alt=\"\" class=\"wp-image-248825\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Which compound matches the IR spectrum? 1,5-hexadiene dipropyl ether hexen-3-ol cyclohexane trans-4-octene 3-hexanol 1-methylcyclopentene The Correct Answer and Explanation is: The correct compound that matches the IR spectrum from the options provided is 1,5-hexadiene. Explanation: To identify the compound that matches an IR spectrum, we need to look for key functional groups and bond vibrations [&hellip;]<\/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-248824","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248824","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=248824"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248824\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248824"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248824"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248824"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}