{"id":234536,"date":"2025-06-14T08:13:38","date_gmt":"2025-06-14T08:13:38","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=234536"},"modified":"2025-06-14T08:13:47","modified_gmt":"2025-06-14T08:13:47","slug":"the-ir-spectrum-of-trans-cinnamic-acid-the-starting-material-for-this-lab-is-given-below","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/14\/the-ir-spectrum-of-trans-cinnamic-acid-the-starting-material-for-this-lab-is-given-below\/","title":{"rendered":"The IR spectrum of trans-cinnamic acid (the starting material for this lab) is given below"},"content":{"rendered":"\n

The IR spectrum of trans-cinnamic acid (the starting material for this lab) is given below. Annotate the spectrum by labeling as many peaks as you can with the appropriate bond type and whether they correspond to stretches or bends. [6 pts] 4000 3000 2000 1500 WAVENUMBER (cm-1) 1000 500 6. Describe the changes that would be observed in a
H NMR spectrum upon conversion of trans- cinnamic acid to 2,3-dibromo-3-phenylpropanoic acid (i.e. dibromocinnamic acid). Make sure to describe which signals would disappear from the starting material and which new signals would appear for the product (e.g. loss of a 3H triplet at 2.5 ppm and appearance of a 2H quartet at 8.5 ppm). [6 pts]<\/p>\n\n\n\n

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

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

Answer to Question 5: IR Spectrum Annotation<\/strong><\/h3>\n\n\n\n

The provided IR spectrum is for trans<\/em>-cinnamic acid (C\u2086H\u2085-CH=CH-COOH). The key absorption peaks are identified below:<\/p>\n\n\n\n

    \n
  • ~3300-2500 cm\u207b\u00b9 (very broad):<\/strong>\u00a0This corresponds to the\u00a0O-H stretch<\/strong>\u00a0of the carboxylic acid functional group. The extreme broadness is due to strong intermolecular hydrogen bonding (dimerization).<\/li>\n\n\n\n
  • ~3060 cm\u207b\u00b9 (sharp, on top of the broad O-H):<\/strong>\u00a0This peak represents the\u00a0sp\u00b2 C-H stretches<\/strong>. This includes both the vinylic (=C-H) and aromatic (Ar-H) C-H bonds.<\/li>\n\n\n\n
  • ~1685 cm\u207b\u00b9 (strong, sharp):<\/strong>\u00a0This is the\u00a0C=O stretch<\/strong>\u00a0(carbonyl) of the carboxylic acid. Its frequency is lowered from the typical ~1710 cm\u207b\u00b9 because it is conjugated with both the alkene and the aromatic ring.<\/li>\n\n\n\n
  • ~1625 cm\u207b\u00b9 (medium):<\/strong>\u00a0This absorption is from the\u00a0C=C stretch<\/strong>\u00a0of the alkene double bond.<\/li>\n\n\n\n
  • ~1580, 1495, 1450 cm\u207b\u00b9 (sharp):<\/strong>\u00a0These peaks are characteristic of\u00a0C=C in-ring stretches<\/strong>\u00a0of the aromatic (phenyl) group.<\/li>\n\n\n\n
  • ~1310 cm\u207b\u00b9 (strong):<\/strong>\u00a0This peak is attributed to the\u00a0C-O stretch<\/strong>\u00a0of the carboxylic acid group.<\/li>\n\n\n\n
  • ~980 cm\u207b\u00b9 (strong, sharp):<\/strong>\u00a0This is a highly diagnostic peak for a\u00a0trans-substituted alkene<\/strong>. It corresponds to the out-of-plane\u00a0=C-H bend<\/strong>.<\/li>\n\n\n\n
  • ~765 cm\u207b\u00b9 and ~690 cm\u207b\u00b9 (strong):<\/strong>\u00a0These two peaks are characteristic out-of-plane\u00a0C-H bends<\/strong>\u00a0for a monosubstituted benzene ring.<\/li>\n<\/ul>\n\n\n\n

    Answer to Question 6: \u00b9H NMR Spectrum Changes<\/strong><\/h3>\n\n\n\n

    The conversion of trans<\/em>-cinnamic acid to 2,3-dibromo-3-phenylpropanoic acid involves the addition of Br\u2082 across the alkene double bond. This chemical transformation leads to significant and predictable changes in the \u00b9H NMR spectrum.<\/p>\n\n\n\n

    Signals that would disappear from the trans<\/em>-cinnamic acid spectrum:<\/strong><\/p>\n\n\n\n

    The most notable change is the disappearance of the signals from the two alkene protons (-CH=CH-). In the starting material, these appear as two distinct 1H doublets<\/strong>. One doublet is typically around 6.4 ppm and the other is further downfield around 7.7 ppm (often overlapping with the aromatic signals). A key feature of these doublets is their large coupling constant (J \u2248 12-18 Hz), which is characteristic of a trans<\/em> configuration. These two signals will be completely absent in the product’s spectrum because the double bond is no longer present.<\/p>\n\n\n\n

    New signals that would appear for the 2,3-dibromo-3-phenylpropanoic acid product:<\/strong><\/p>\n\n\n\n

    In place of the alkene signals, new signals corresponding to the two protons on the newly formed single bond (-CH(Br)-CH(Br)-) will emerge. These two protons are in different chemical environments and will couple to each other, resulting in two new 1H doublets<\/strong>. These signals will appear in the aliphatic region, typically between 4.5 and 5.5 ppm, due to the deshielding effect of the adjacent bromine atoms and other functional groups. The coupling constant for these new doublets will be much smaller (J \u2248 2-10 Hz) than the one observed for the trans<\/em>-alkene protons.<\/p>\n\n\n\n

    The signals for the 5H aromatic protons<\/strong> (multiplet, ~7.4 ppm) and the 1H carboxylic acid proton<\/strong> (broad singlet, >10 ppm) will remain in the spectrum, although their exact chemical shifts might be slightly altered due to the change in the molecule’s electronic structure.<\/p>\n\n\n\n

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

    The IR spectrum of trans-cinnamic acid (the starting material for this lab) is given below. Annotate the spectrum by labeling as many peaks as you can with the appropriate bond type and whether they correspond to stretches or bends. [6 pts] 4000 3000 2000 1500 WAVENUMBER (cm-1) 1000 500 6. Describe the changes that would […]<\/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-234536","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/234536","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=234536"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/234536\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=234536"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=234536"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=234536"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}