{"id":245342,"date":"2025-07-06T04:05:30","date_gmt":"2025-07-06T04:05:30","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=245342"},"modified":"2025-07-06T04:05:33","modified_gmt":"2025-07-06T04:05:33","slug":"in-the-ir-data-provided-assign-the-major-peaks-and-identify-which-spectrum-is-for-9-fluorenone-and-which-spectrum-is-9-fluorenol","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/06\/in-the-ir-data-provided-assign-the-major-peaks-and-identify-which-spectrum-is-for-9-fluorenone-and-which-spectrum-is-9-fluorenol\/","title":{"rendered":"In the IR data provided, assign the major peaks and identify which spectrum is for 9-fluorenone and which spectrum is 9-fluorenol"},"content":{"rendered":"\n<p>In the IR data provided, assign the major peaks and identify which spectrum is for 9-fluorenone and which spectrum is 9-fluorenol. Drag the answers to the appropriate designated box on the image. If an item is not used, drag it to the &#8220;if not used, drag here&#8221; box at the bottom of the image. If needed, you can use an option more than once. O-H Molecules to assign: Ident 9-fluorenol C-H O-H C=O If not N-H C=C O-H N-H OH Identi 9-fluorenone C=C C=O C=O C-H N<\/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\/image-217.png\" alt=\"\" class=\"wp-image-245343\"\/><\/figure>\n\n\n\n<p><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>Based on the principles of IR spectroscopy, here is the correct assignment for the given spectra and molecules.<\/p>\n\n\n\n<p><strong>Left Spectrum:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Molecule Identification:<\/strong>\u00a0<strong>9-fluorenone<\/strong><\/li>\n\n\n\n<li><strong>Peak Assignments:<\/strong>\n<ul class=\"wp-block-list\">\n<li><strong>C-H:<\/strong>\u00a0The box points to the sharp peaks just above 3000 cm\u207b\u00b9, which correspond to the aromatic C-H stretches.<\/li>\n\n\n\n<li><strong>C=O:<\/strong>\u00a0The box points to the very strong, sharp absorption at approximately 1715 cm\u207b\u00b9. This is the characteristic peak for a ketone&#8217;s carbonyl (C=O) group. The conjugation with the aromatic rings lowers the frequency slightly from a simple aliphatic ketone.<\/li>\n\n\n\n<li>The box pointing to the baseline around 3300 cm\u207b\u00b9 should not have a label, as there is no O-H peak present.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p><strong>Right Spectrum:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Molecule Identification:<\/strong>\u00a0<strong>9-fluorenol<\/strong><\/li>\n\n\n\n<li><strong>Peak Assignments:<\/strong>\n<ul class=\"wp-block-list\">\n<li><strong>O-H:<\/strong>\u00a0The box points to the very broad, strong peak centered around 3300 cm\u207b\u00b9. This is the classic signature of an alcohol&#8217;s O-H stretching vibration.<\/li>\n\n\n\n<li><strong>C=C:<\/strong>\u00a0The box points to the absorptions in the 1450-1600 cm\u207b\u00b9 region, which are characteristic of carbon-carbon double bond stretching within the aromatic rings.<\/li>\n\n\n\n<li>The box pointing to the baseline around 1700 cm\u207b\u00b9 should not have a label, as the absence of a peak here confirms the C=O group is not present.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p><strong>Unused Labels:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>N-H:<\/strong>\u00a0Neither 9-fluorenone nor 9-fluorenol contains a nitrogen-hydrogen bond, so this label is not used.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation<\/strong><\/h3>\n\n\n\n<p>The key to distinguishing between the IR spectra of 9-fluorenone and 9-fluorenol is to look for the characteristic peaks of their different functional groups: a ketone (C=O) for 9-fluorenone and an alcohol (O-H) for 9-fluorenol.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Identifying 9-fluorenol:<\/strong>\u00a0The spectrum on the right displays a very prominent, broad, and strong absorption band in the region of 3200-3600 cm\u207b\u00b9. This is the definitive signal for the O-H stretching vibration of an alcohol. The broadness is due to hydrogen bonding between the alcohol molecules. This spectrum also lacks a strong peak around 1700 cm\u207b\u00b9, confirming the absence of a carbonyl group. Therefore, the right spectrum belongs to\u00a0<strong>9-fluorenol<\/strong>. The peaks around 1600 cm\u207b\u00b9 are correctly assigned to the C=C bonds of the aromatic rings.<\/li>\n\n\n\n<li><strong>Identifying 9-fluorenone:<\/strong>\u00a0The spectrum on the left lacks the broad O-H peak seen in the other spectrum. Instead, it shows a very strong, sharp absorption peak at approximately 1715 cm\u207b\u00b9. This is the characteristic position for a C=O (carbonyl) stretch in a conjugated ketone. The peaks just above 3000 cm\u207b\u00b9 are due to the C-H stretching of the aromatic rings. The absence of the O-H peak and the presence of the strong C=O peak definitively identify the left spectrum as belonging to\u00a0<strong>9-fluorenone<\/strong>.<\/li>\n\n\n\n<li><strong>Unused Label:<\/strong>\u00a0Both molecules are hydrocarbons with an oxygen atom, but neither contains nitrogen. Thus, the N-H label is not applicable to either spectrum and should be placed in the &#8220;if not used&#8221; category.<\/li>\n<\/ol>\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-banner5-633.jpeg\" alt=\"\" class=\"wp-image-245344\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>In the IR data provided, assign the major peaks and identify which spectrum is for 9-fluorenone and which spectrum is 9-fluorenol. Drag the answers to the appropriate designated box on the image. If an item is not used, drag it to the &#8220;if not used, drag here&#8221; box at the bottom of the image. If [&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-245342","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245342","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=245342"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245342\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=245342"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=245342"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=245342"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}