To determine which compound matches the IR spectrum, we need to analyze the characteristic IR absorption bands of each compound and compare them with typical functional group absorptions.<\/p>\n\n\n\n
- \n
- Cyclohexane<\/strong>: This is a saturated hydrocarbon, and its IR spectrum would show peaks around 2900 cm\u207b\u00b9 for C-H stretching and 1450 cm\u207b\u00b9 for C-H bending. There would be no significant peaks for functional groups like alcohols, alkenes, or ethers.<\/li>\n\n\n\n
- Dipropyl Ether<\/strong>: Dipropyl ether has an ether group (-O-) and alkyl chains. In its IR spectrum, you would expect a strong absorption around 1100-1200 cm\u207b\u00b9 for the C-O stretch. There will also be C-H stretches around 2900 cm\u207b\u00b9 and C-H bending around 1450 cm\u207b\u00b9, but no significant peaks for alcohols or alkenes.<\/li>\n\n\n\n
- Trans-4-Octene<\/strong>: Trans-4-octene contains a C=C double bond and will show a strong absorption around 1600 cm\u207b\u00b9 for the C=C stretch. In addition, there will be C-H stretching around 2900 cm\u207b\u00b9 and bending around 1450 cm\u207b\u00b9. The trans-configuration can also affect the exact position of these bands.<\/li>\n\n\n\n
- 1,5-Hexadiene<\/strong>: This compound has two double bonds (C=C) in the chain, so its IR spectrum will show the typical C=C stretching absorption around 1600 cm\u207b\u00b9. It will also show C-H stretches around 2900 cm\u207b\u00b9 and bending around 1450 cm\u207b\u00b9. The two conjugated alkenes might lead to some slight shifts in the C=C stretching band.<\/li>\n\n\n\n
- 1-Methylcyclopentene<\/strong>: This compound contains a C=C double bond and a methyl group. The C=C stretching absorption will appear around 1600 cm\u207b\u00b9, and C-H stretches will appear near 2900 cm\u207b\u00b9. There could also be a distinct peak for the methyl group, but no significant oxygen-related bands like those found in alcohols or ethers.<\/li>\n\n\n\n
- 3-Hexanol<\/strong>: This is an alcohol, and its IR spectrum will show a strong, broad O-H stretch around 3300 cm\u207b\u00b9, which is a key identifier for alcohols. Additionally, there will be C-H stretching around 2900 cm\u207b\u00b9 and C-H bending around 1450 cm\u207b\u00b9, but no C=C stretch.<\/li>\n\n\n\n
- 1-Hexen-3-ol<\/strong>: This compound contains both an alcohol (-OH) group and an alkene (C=C) bond. The IR spectrum would show a broad O-H stretch around 3300 cm\u207b\u00b9, along with a C=C stretch around 1600 cm\u207b\u00b9. It will also show C-H stretches and bending in the usual regions.<\/li>\n<\/ol>\n\n\n\n
Conclusion:<\/h3>\n\n\n\n
Based on these general characteristics:<\/p>\n\n\n\n
- \n
- 3-Hexanol<\/strong> and 1-Hexen-3-ol<\/strong> would show the distinct O-H stretch, with 1-Hexen-3-ol also showing the C=C stretch.<\/li>\n\n\n\n
- Trans-4-Octene<\/strong> and 1,5-Hexadiene<\/strong> would show prominent C=C stretches.<\/li>\n\n\n\n
- Dipropyl ether<\/strong> and Cyclohexane<\/strong> would show only C-H stretches with no significant functional group absorptions.<\/li>\n<\/ul>\n\n\n\n
Without the actual IR spectrum data, the most likely matches based on functional group analysis are 3-Hexanol<\/strong> for alcohols and 1-Hexen-3-ol<\/strong> for alcohols with a C=C bond. If a C=C stretch is observed, 1-Hexen-3-ol<\/strong> would be a better match.<\/p>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner8-324.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"Which compound matches the IR spectrum? cyclohexane dipropyl ether trans-4-octene 1,5-hexadiene 1-methylcyclopentene 3-hexanol 1-hexen-3-ol The Correct Answer and Explanation is: To determine which compound matches the IR spectrum, we need to analyze the characteristic IR absorption bands of each compound and compare them with typical functional group absorptions. Conclusion: Based on these general characteristics: Without […]<\/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-248819","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248819","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=248819"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248819\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248819"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248819"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248819"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- Trans-4-Octene<\/strong> and 1,5-Hexadiene<\/strong> would show prominent C=C stretches.<\/li>\n\n\n\n
- Dipropyl Ether<\/strong>: Dipropyl ether has an ether group (-O-) and alkyl chains. In its IR spectrum, you would expect a strong absorption around 1100-1200 cm\u207b\u00b9 for the C-O stretch. There will also be C-H stretches around 2900 cm\u207b\u00b9 and C-H bending around 1450 cm\u207b\u00b9, but no significant peaks for alcohols or alkenes.<\/li>\n\n\n\n