{"id":252873,"date":"2025-07-11T14:19:56","date_gmt":"2025-07-11T14:19:56","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=252873"},"modified":"2025-07-11T14:19:58","modified_gmt":"2025-07-11T14:19:58","slug":"interpretation-of-ir-spectra-of-vanillin-and-vanillyl-alcohol","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/11\/interpretation-of-ir-spectra-of-vanillin-and-vanillyl-alcohol\/","title":{"rendered":"Interpretation of IR spectra of vanillin and vanillyl alcohol."},"content":{"rendered":"\n

Interpretation of IR spectra of vanillin and vanillyl alcohol. For vanillin and vanillyl alcohol, predict the IR signal using the Handout provided. Vanillin Predict IR signal in Wavenumber, this compound cm (yin) H Bond types O-H (phenol) O-H (alcohol) C-H (sp3) on alkane C-H (sp2) on the ring C-H (sp2) on alkene C-H in the aldehyde group C=O in the aldehyde group C=C in the ring C-O in ether, ester, alcohol CH3 OH Vanillyl alcohol Predict IR signal in Wavenumber, this compound cm (yin) Bond types O-H (phenol) O-H (alcohol) C-H (sp3) on alkane C-H (sp2) on the ring C-H (sp2) on alkene C-H in the aldehyde group C=O in the aldehyde group C=C in the ring C-O in ether, ester, alcohol CH3 OH<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Here are the completed tables for vanillin and vanillyl alcohol with the predicted IR signals and a detailed explanation.<\/p>\n\n\n\n

A. Vanillin<\/strong><\/h3>\n\n\n\n
#<\/td>Bond types<\/td>Wavenumber, cm\u207b\u00b9<\/td>Predict IR signal in this compound (Y\/N)<\/td><\/tr>
1<\/td>O\u2013H (phenol)<\/td>3550\u20133200 (broad)<\/td>Y<\/td><\/tr>
2<\/td>O\u2013H (alcohol)<\/td><\/td>N<\/td><\/tr>
3<\/td>C\u2013H (sp\u00b3) on alkane<\/td>2950\u20132850<\/td>Y<\/td><\/tr>
4<\/td>C\u2013H (sp\u00b2) on the ring<\/td>3100\u20133000<\/td>Y<\/td><\/tr>
5<\/td>C\u2013H (sp\u00b2) on alkene<\/td><\/td>N<\/td><\/tr>
6<\/td>C\u2013H in the aldehyde group<\/td>2850\u20132700 (two peaks)<\/td>Y<\/td><\/tr>
7<\/td>C=O in the aldehyde group<\/td>1700\u20131680 (strong)<\/td>Y<\/td><\/tr>
8<\/td>C=C in the ring<\/td>1600\u20131450<\/td>Y<\/td><\/tr>
9<\/td>C\u2013O in ether, ester, alcohol<\/td>1300\u20131000<\/td>Y<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

B. Vanillyl alcohol<\/strong><\/h3>\n\n\n\n
#<\/td>Bond types<\/td>Wavenumber, cm\u207b\u00b9<\/td>Predict IR signal in this compound (Y\/N)<\/td><\/tr>
1<\/td>O\u2013H (phenol)<\/td>3550\u20133200 (very broad)<\/td>Y<\/td><\/tr>
2<\/td>O\u2013H (alcohol)<\/td>3650\u20133200 (very broad)<\/td>Y<\/td><\/tr>
3<\/td>C\u2013H (sp\u00b3) on alkane<\/td>2950\u20132850<\/td>Y<\/td><\/tr>
4<\/td>C\u2013H (sp\u00b2) on the ring<\/td>3100\u20133000<\/td>Y<\/td><\/tr>
5<\/td>C\u2013H (sp\u00b2) on alkene<\/td><\/td>N<\/td><\/tr>
6<\/td>C\u2013H in the aldehyde group<\/td><\/td>N<\/td><\/tr>
7<\/td>C=O in the aldehyde group<\/td><\/td>N<\/td><\/tr>
8<\/td>C=C in the ring<\/td>1600\u20131450<\/td>Y<\/td><\/tr>
9<\/td>C\u2013O in ether, ester, alcohol<\/td>1300\u20131000<\/td>Y<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Explanation<\/strong><\/h3>\n\n\n\n

The predictions for the IR spectra of vanillin and vanillyl alcohol are based on the functional groups present in each molecule.<\/p>\n\n\n\n

Vanillin Analysis:<\/strong>
Vanillin’s structure contains a phenolic hydroxyl (O-H) group, an aldehyde (CHO), an ether (O-CH3), and an aromatic ring. The phenolic O\u2013H<\/strong> bond gives a characteristic broad signal around 3550\u20133200 cm\u207b\u00b9. The methyl ether group has sp\u00b3 C\u2013H<\/strong> bonds, which produce stretches just below 3000 cm\u207b\u00b9 (2950\u20132850 cm\u207b\u00b9). The aromatic ring shows sp\u00b2 C\u2013H<\/strong> stretches just above 3000 cm\u207b\u00b9 (3100\u20133000 cm\u207b\u00b9) and C=C<\/strong> bond stretches in the 1600\u20131450 cm\u207b\u00b9 region. The most diagnostic features of vanillin are from its aldehyde group: a strong, sharp C=O<\/strong> (carbonyl) stretch, which is conjugated to the aromatic ring and appears at a lower frequency around 1700\u20131680 cm\u207b\u00b9, and two weak but characteristic aldehyde C\u2013H<\/strong> stretches between 2850\u20132700 cm\u207b\u00b9. Finally, the C\u2013O<\/strong> bonds of the ether and phenol groups will show strong absorptions in the fingerprint region (1300\u20131000 cm\u207b\u00b9).<\/p>\n\n\n\n

Vanillyl Alcohol Analysis:<\/strong>
In vanillyl alcohol, the aldehyde group has been reduced to a primary alcohol (CH2OH). It still has the phenolic O\u2013H, ether, and aromatic ring, so it will share many signals with vanillin, including the sp\u00b3 C\u2013H, sp\u00b2 C\u2013H, aromatic C=C, and C\u2013O stretches. However, there are critical differences. Vanillyl alcohol has two types of hydroxyl groups (phenolic and alcoholic), which will result in a very broad and intense O\u2013H<\/strong> signal spanning the 3650\u20133200 cm\u207b\u00b9 range. Most importantly, vanillyl alcohol lacks<\/strong> an aldehyde group. Therefore, its IR spectrum will be missing the strong C=O<\/strong> carbonyl peak around 1700 cm\u207b\u00b9 and the specific aldehyde C\u2013H<\/strong> stretches. This absence is the clearest indicator of the conversion from vanillin to vanillyl alcohol.<\/p>\n\n\n\n

\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

Interpretation of IR spectra of vanillin and vanillyl alcohol. For vanillin and vanillyl alcohol, predict the IR signal using the Handout provided. Vanillin Predict IR signal in Wavenumber, this compound cm (yin) H Bond types O-H (phenol) O-H (alcohol) C-H (sp3) on alkane C-H (sp2) on the ring C-H (sp2) on alkene C-H in the […]<\/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-252873","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/252873","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=252873"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/252873\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=252873"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=252873"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=252873"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}