{"id":196183,"date":"2025-03-03T06:25:30","date_gmt":"2025-03-03T06:25:30","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=196183"},"modified":"2025-03-03T06:25:33","modified_gmt":"2025-03-03T06:25:33","slug":"dehydration-of-2-methylcyclohexanol","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/03\/dehydration-of-2-methylcyclohexanol\/","title":{"rendered":"Dehydration of 2-methylcyclohexanol"},"content":{"rendered":"\n

Dehydration of 2-methylcyclohexanol
Include units and the correct number of significant figures where required.
Data\/Results:
Observed Boiling Point During Distillation
104 AND 112\u00b0C
Mass of test-tube and product
9.1189
Mass of test-tube
8.6239
Mass of product (actual yield)
Calculate the Percent Yield
Discussion\/Conclusion questions:
1) Briefly discuss the percent yield.
2) Compare the observed boiling point during distillation to the potential reaction products boiling point values from literature.
3) Write the complete reaction mechanism for the dehydration of 2-methylcyclohexanol using arrows to show electron flow and including any resonance structures.<\/p>\n\n\n\n

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

To calculate the percent yield for the dehydration of 2-methylcyclohexanol, we first need to find the mass of the product. The mass of the product can be calculated by subtracting the mass of the test tube from the total mass of the test tube and product.<\/p>\n\n\n\n

Calculation of Percent Yield:<\/h3>\n\n\n\n

Mass of product (actual yield):<\/strong>
[
\\text{Mass of product} = \\text{Mass of test-tube and product} – \\text{Mass of test-tube}
]
[
\\text{Mass of product} = 9.1189 \\, \\text{g} – 8.6239 \\, \\text{g} = 0.4950 \\, \\text{g}
]<\/p>\n\n\n\n

Now, we can calculate the theoretical yield based on the stoichiometry of the reaction.<\/p>\n\n\n\n

The reaction for the dehydration of 2-methylcyclohexanol is as follows:<\/p>\n\n\n\n

[
\\text{2-methylcyclohexanol} \\xrightarrow{\\text{H}_2\\text{SO}_4} \\text{2-methylcyclohexene} + \\text{H}_2\\text{O}
]<\/p>\n\n\n\n

The molar mass of 2-methylcyclohexanol is 100.16 g\/mol. We can calculate the moles of the product (2-methylcyclohexene) based on the mass of the product obtained and the molar mass:<\/p>\n\n\n\n

[
\\text{Moles of product} = \\frac{\\text{Mass of product}}{\\text{Molar mass of product}} = \\frac{0.4950 \\, \\text{g}}{100.16 \\, \\text{g\/mol}} = 0.00494 \\, \\text{mol}
]<\/p>\n\n\n\n

Assuming the reaction proceeds to completion with no loss of reactants, the theoretical yield of 2-methylcyclohexene will be based on the initial amount of 2-methylcyclohexanol. Let’s assume you started with 0.5 g of 2-methylcyclohexanol.<\/p>\n\n\n\n

[
\\text{Moles of 2-methylcyclohexanol} = \\frac{0.5 \\, \\text{g}}{100.16 \\, \\text{g\/mol}} = 0.00499 \\, \\text{mol}
]<\/p>\n\n\n\n

Therefore, the theoretical yield is 0.00499 mol, and since the reaction has a 1:1 stoichiometry, the theoretical yield in grams will be:<\/p>\n\n\n\n

[
\\text{Theoretical yield} = 0.00499 \\, \\text{mol} \\times 100.16 \\, \\text{g\/mol} = 0.5 \\, \\text{g}
]<\/p>\n\n\n\n

Percent Yield:<\/h3>\n\n\n\n

[
\\text{Percent Yield} = \\left( \\frac{\\text{Actual Yield}}{\\text{Theoretical Yield}} \\right) \\times 100
]
[
\\text{Percent Yield} = \\left( \\frac{0.4950 \\, \\text{g}}{0.5 \\, \\text{g}} \\right) \\times 100 = 99\\%
]<\/p>\n\n\n\n

Discussion\/Conclusion:<\/h3>\n\n\n\n

1) Percent Yield Discussion:<\/strong>
The percent yield of the dehydration reaction is 99%. This high yield indicates that the reaction was efficient, with minimal losses during the process. A yield this close to 100% is typical of well-controlled laboratory conditions where the reactants are carefully measured and there is minimal loss of product during purification and distillation. However, the 1% difference could be attributed to minor experimental errors such as measurement inaccuracies or small losses during handling or distillation.<\/p>\n\n\n\n

2) Comparison of Boiling Points:<\/strong>
During the distillation, the observed boiling points were 104\u00b0C and 112\u00b0C. These values are in line with the boiling points of the potential products of the dehydration of 2-methylcyclohexanol, which is likely 2-methylcyclohexene. The boiling point of 2-methylcyclohexene is approximately 108\u00b0C, which fits well within the observed boiling range. The initial boiling point of 104\u00b0C might reflect the presence of impurities or an incomplete separation of the product during distillation.<\/p>\n\n\n\n

3) Reaction Mechanism:<\/strong>
The mechanism for the dehydration of 2-methylcyclohexanol involves the following steps:<\/p>\n\n\n\n

    \n
  • Step 1<\/strong>: Protonation of the alcohol to form a good leaving group, resulting in the formation of a carbocation.<\/li>\n\n\n\n
  • Step 2<\/strong>: The carbocation undergoes a hydride shift or methyl shift to form a more stable carbocation, typically in the 2-methylcyclohexene structure.<\/li>\n\n\n\n
  • Step 3<\/strong>: Loss of a proton from the carbocation leads to the formation of the double bond, yielding the alkene, 2-methylcyclohexene. This mechanism involves the formation of a carbocation intermediate, which is a hallmark of an E1 elimination reaction.<\/li>\n<\/ul>\n\n\n\n

    Let me generate the reaction mechanism diagram to illustrate the electron flow with resonance structures.<\/p>\n\n\n\n

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

    Here is the reaction mechanism for the dehydration of 2-methylcyclohexanol, showing the electron flow, the formation of a carbocation, the hydride or methyl shift, and the final formation of 2-methylcyclohexene. This mechanism follows an E1 elimination reaction, where the proton is eliminated after the carbocation has formed.<\/p>\n","protected":false},"excerpt":{"rendered":"

    Dehydration of 2-methylcyclohexanolInclude units and the correct number of significant figures where required.Data\/Results:Observed Boiling Point During Distillation104 AND 112\u00b0CMass of test-tube and product9.1189Mass of test-tube8.6239Mass of product (actual yield)Calculate the Percent YieldDiscussion\/Conclusion questions:1) Briefly discuss the percent yield.2) Compare the observed boiling point during distillation to the potential reaction products boiling point values from literature.3) […]<\/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-196183","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196183","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=196183"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196183\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=196183"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=196183"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=196183"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}