Q69E - Organic Chemistry (Mcmurry)

Question

Fill in the missing reagents a–h in the following scheme:

Step-by-Step Solution

Answer

The missing reagents in the above reaction are:

1. $LiAl{H}_4, H_3{O}^{+}$
2. $POC{l}_3$, pyridine
3. $KMn{O}_4, H_3{O}^{+}$
   
4. $C{H}_{3}C{H}_{2}OH, H^{+}$
   
5. $N{a}^{+}OE{t}^{-}$
   
6. $H_3{O}^{+}$
   
7. $N{a}^{+}OE{t}^{-}, C{H}_{3}Br$
   
8. $H_3{O}^{+}$, heat

Explanation of the process

The reactions in the given scheme are

1. Reduction of the ketone to alcohol;
2. Dehydration of the alcohol to alkene;
3. Ring cleavage by oxidation;
4. Esterification of the carboxylic acids;
5. The esters undergo Dieckmann cyclization;
6. The esters undergo Dieckmann cyclization;
7. The methyl group substituted;
8. Lastly, hydrolysis and decarboxylation.

Missing reagents (a-h) in the given scheme

1. $LiAl{H}_4, H_3{O}^{+}$
2. $POC{l}_3$, pyridine
3. $KMn{O}_4, H_3{O}^{+}$
   
4. $C{H}_{3}C{H}_{2}OH, H^{+}$
   
5. $N{a}^{+}OE{t}^{-}$
   
6. $H_3{O}^{+}$
   
7. $N{a}^{+}OE{t}^{-}, C{H}_{3}Br$
   
8. $H_3{O}^{+}, heat$

Explanation of the steps

1. Reduction with$LiAl{H}_4$, $H_3{O}^{+}$, cyclohexanone reduced to cyclohexanol.
2. Treated with$POC{l}_3$, pyridine, cyclohexanol converted into cyclohexene by dehydration.
3. Oxidation with$KMn{O}_4$, $H_3{O}^{+}$, the cyclohexene undergoes cleavage and converted into dicarboxylic acid.
4. Treating with ethanol in presence of acid, the dicarboxylic acid converts into diester.
5. Treating with$N{a}^{+}OE{t}^{-}$, $H_3{O}^{+}$, the diester converts into β-keto ester by Dieckmann cyclization.
6. Treating with $N{a}^{+}OE{t}^{-}$, $H_3{O}^{+}$,  the diester converts into β-keto ester by Dieckmann cyclization.
7. Treating with methyl bromide in presence of sodium ethoxide, a methyl group gets substituted of the ester.
8. Treating with diluted acid, the ester converts into keto acid by hydrolysis; and finally forms 2-methylcyclopentanone on heating by decarboxylation.

Conclusion

The full scheme is like