Question

Draw the products formed when $\textbf{A}$ or $\textbf{b}$ is treated with each reagent. In some cases, no reaction occurs. a. NaBH$4$, CH$_3$OH b. [1] LiAIH$_4$; [2] H$_2$O c. [1] CH$_3$MgBr (excess); [2] H$_2$O d. [1] C$_6$H$_5$Li (excess); [2] H$_2$O e. Na$_2$Cr$_2$O$_7$, H$_2$SO$_4$, H$_2$O

   Draw the products formed when $\textbf{A}$ or $\textbf{b}$ is treated with each reagent. In some cases, no reaction occurs.
a. NaBH$4$, CH$_3$OH
b. [1] LiAIH$_4$; [2] H$_2$O
c. [1] CH$_3$MgBr (excess); [2] H$_2$O
d. [1] C$_6$H$_5$Li (excess); [2] H$_2$O
e. Na$_2$Cr$_2$O$_7$, H$_2$SO$_4$, H$_2$O

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Organic Chemistry

Organic Chemistry

Janice Gorzynski… 4th Edition

Chapter 20, Problem 36

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Solved by Expert Colton K

06/01/2019

Step 1

This is a reducing agent that can reduce carbonyl groups to alcohols. However, compound B is an ester, which cannot be reduced by this reagent. Therefore, no reaction occurs for compound B. For compound A, the carbonyl group is reduced to a secondary alcohol. Show more…

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Draw the products formed when $\textbf{A}$ or $\textbf{b}$ is treated with each reagent. In some cases, no reaction occurs. a. NaBH$4$, CH$_3$OH b. [1] LiAIH$_4$; [2] H$_2$O c. [1] CH$_3$MgBr (excess); [2] H$_2$O d. [1] C$_6$H$_5$Li (excess); [2] H$_2$O e. Na$_2$Cr$_2$O$_7$, H$_2$SO$_4$, H$_2$O

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Transcript

00:01 Okay, in this problem, we're starting with this ketone and subjecting it to a variety of reaction conditions.

00:09 The first being sodium borohydride and methanol.

00:15 Of course, for this one, we imagine this is just the hydride, attacking the carbonyl directly.

00:24 Then we'll generate this secondary alcohol.

00:30 Second condition, they would like you to use lithium, aluminum, hydropy, water.

00:38 Actually, this reactivity is the exact same.

00:41 It's just a slightly more reducing condition.

00:44 So it's also just an h -minus precursor, which will attack directly and get you to the same product.

00:53 For c, they want you to use a methyl greenyard, followed by water.

01:06 This methyl greenyard will attack directly, and you'll generate this tertiary alcohol.

01:18 For part d, kind of running out of spacing, let's see how we just redraw it here.

01:28 Okay, for d, we're using this phenolithium reagent and then water.

01:40 And of course, this is just going to act as your carbon nucleophile, generate this tertiary alcohol.

01:54 And finally, they would like to subject it to some oxidizing conditions in chromium six species.

02:03 And this is actually already the most oxidized condition, so nothing will happen in this case.

02:15 For b, we actually have a substituted cyclohexane...

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