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Chapter 4: Problem 146

The stable hemiacetals is/are (A) CC1(O)CCCCC1 (B) OC1CCCCO1 (C) \(\mathrm{Br}_{3} \mathrm{C} \quad \mathrm{OCH}_{3}\) (D)

Short Answer

Expert verified
The stable hemiacetal is (A) with the structure CC1(O)CCCCC1, as it meets the criteria of having an -OH group, an -OR group, and two other groups in a six-membered ring.

Step by step solution

01

Structure of molecule (A)

For molecule (A), the SMILES notation is CC1(O)CCCCC1. This tells us that the molecule has a six-membered ring with an oxygen atom in the ring, and a methyl group attached to one of the carbons. The structure meets the criteria for being a stable hemiacetal since it has an -OH group, an -OR group, and two other groups in a six-membered ring.
02

Structure of molecule (B)

For molecule (B), the SMILES notation is OC1CCCCO1. The molecule has a five-membered ring with two oxygen atoms in the ring. This structure does not meet the criteria for being a hemiacetal as no carbon atom in the ring has an -OH group and an -OR group within the same ring.
03

Structure of molecule (C)

For molecule (C), the formula is \(\mathrm{Br}_{3} \mathrm{C} \quad \mathrm{OCH}_{3}\). This molecule does not have a ring structure and therefore cannot be a stable hemiacetal.
04

Conclusion

Among the given molecules, only molecule (A) meets the criteria for being a stable hemiacetal. Therefore, the correct answer is (A).

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Most popular questions from this chapter

consider the following sequence of reactions. CC(C)C(=O)CN (A) $\frac{\text { (i) } \mathrm{LiAlH}_{4}}{\text { (ii) } \mathrm{H}_{2} \mathrm{O}}{\longrightarrow}(\mathrm{B})$ The final product (B) is (A) CNC(C)(O)C(C)C (B) [X][M]C(C)C(C)C (C) CC(C)NC(C)C(C)C (D) CN=C(C)C(C)Br

Consider the following aldehydes with respect to their reactivity toward addition reaction with a given nucleoppile (I) O=Cc1ccccc1 (II) O=Cc1ccc([N+](=O)[O-])cc1 (III) (IV) N#Cc1cc([N+](=O)[O-])ccc1C=O The order of reactivity is (A) III > I > II > IV (B) IV > II > I > III (C) II > IV > I > III (D) \(\mathrm{IV}>\mathrm{III}>\mathrm{II}>\mathrm{I}\)

The order of electrophilic substitution reaction in the following compound is (I) Anisole (II) Acetophenone (III) Phenylethanoate (IV) Benzene (A) \(\mathrm{I}>\mathrm{II}>\mathrm{III}>\mathrm{IV}\) (B) \(\mathrm{I}>\mathrm{III}>\mathrm{II}>\mathrm{IV}\) \((\mathrm{C}) \mathrm{I}>\mathrm{III}>\mathrm{IV}>\mathrm{II}\) (D) II > I > IV > III

$$ \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3} \longrightarrow \mathrm{CH}_{3}-\mathrm{CHO} $$ ' \(\mathrm{X}\) ' can be : (A) \(\mathrm{O}_{3}, \mathrm{Zn}\) (B) \(\mathrm{KMnO}_{4} / \mathrm{OH}, \Delta\) (C) (i) \(\mathrm{HCO}_{3} \mathrm{H}\), (ii) \(\mathrm{HIO}_{4}\) (D) Both (A) and (C)

How many stereoisomers are possible for (Q)? $$ \bigcap_{\mathrm{O}} / \underset{\mathrm{HCN}}{\stackrel{\mathrm{NaCN}}{\longrightarrow}}(\mathrm{P}) \frac{(\mathrm{i}) \mathrm{LiAlH}_{4}}{(\mathrm{ii}) \mathrm{H}_{2} \mathrm{O}}(\mathrm{Q}) $$ (A) 2 (B) 4 (C) 1 (D) 3
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