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Which of the follow reactions would produce a primary alcohol? Hold and drag to reorder 1. No 2. = Yes 3. No = No = Yes = Yes 1) NABH, CH,OH 2) 1. LIAIH,, ether 2. H,0" 1. excess LIAIH,, ether 2. Н,о NABH, CH,OH но, 4.

Which of the follow reactions would produce a primary alcohol? Hold and drag to reorder 1. No 2. = Yes 3. No = No = Yes = Yes 1) NABH, CH,OH 2) 1. LIAIH,, ether 2. H,0" 1. excess LIAIH,, ether 2. Н,о NABH, CH,OH но, 4.

Chemistry
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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### H14.11 - Level 2 **Status:** Unanswered • 3 attempts left #### Question: Which of the following reactions would produce a primary alcohol? **Instructions:** Hold and drag to reorder | | Answer | |---|---------| | 1 | No | | 2 | Yes | | 3 | No | | 4 | No | | 5 | Yes | | 6 | Yes | #### Reaction Diagrams: 1) The first reaction involves a ketone treated with NaBH₄ in methanol (CH₃OH). **Diagram:** - Reactant: A ketone with a four-carbon chain. - Reagent: Sodium borohydride (NaBH₄) in methanol. - Product: The ketone is reduced to a secondary alcohol (OH group on the second carbon). 2) The second reaction involves an aldehyde treated first with LiAlH₄ in ether, followed by water (H₂O). **Diagram:** - Reactant: An aldehyde with a three-carbon chain. - Reagent: 1. Lithium aluminium hydride (LiAlH₄) in ether. 2. Water (H₂O). - Product: The aldehyde is reduced to a primary alcohol (OH group on the first carbon). 3) The third reaction involves a cyclic ester (lactone) treated with excess LiAlH₄ in ether, followed by water (H₂O). **Diagram:** - Reactant: A five-membered cyclic ester. - Reagent: 1. Excess lithium aluminium hydride (LiAlH₄) in ether. 2. Water (H₂O). - Product: The ester is reduced to a diol (two OH groups on adjacent carbons), transforming the ring into an open-chain diol. 4) The fourth reaction involves a ketone treated with NaBH₄ in methanol. **Diagram:** - Reactant: A ketone with a four-carbon chain and a hydroxyl (OH) group at the fourth carbon. - Reagent: Sodium borohydride (NaBH₄) in methanol. - Product: The ketone is reduced to a secondary alcohol (OH group on the second carbon), keeping the additional OH group unchanged. **Summary
Transcribed Image Text:### H14.11 - Level 2 **Status:** Unanswered • 3 attempts left #### Question: Which of the following reactions would produce a primary alcohol? **Instructions:** Hold and drag to reorder | | Answer | |---|---------| | 1 | No | | 2 | Yes | | 3 | No | | 4 | No | | 5 | Yes | | 6 | Yes | #### Reaction Diagrams: 1) The first reaction involves a ketone treated with NaBH₄ in methanol (CH₃OH). **Diagram:** - Reactant: A ketone with a four-carbon chain. - Reagent: Sodium borohydride (NaBH₄) in methanol. - Product: The ketone is reduced to a secondary alcohol (OH group on the second carbon). 2) The second reaction involves an aldehyde treated first with LiAlH₄ in ether, followed by water (H₂O). **Diagram:** - Reactant: An aldehyde with a three-carbon chain. - Reagent: 1. Lithium aluminium hydride (LiAlH₄) in ether. 2. Water (H₂O). - Product: The aldehyde is reduced to a primary alcohol (OH group on the first carbon). 3) The third reaction involves a cyclic ester (lactone) treated with excess LiAlH₄ in ether, followed by water (H₂O). **Diagram:** - Reactant: A five-membered cyclic ester. - Reagent: 1. Excess lithium aluminium hydride (LiAlH₄) in ether. 2. Water (H₂O). - Product: The ester is reduced to a diol (two OH groups on adjacent carbons), transforming the ring into an open-chain diol. 4) The fourth reaction involves a ketone treated with NaBH₄ in methanol. **Diagram:** - Reactant: A ketone with a four-carbon chain and a hydroxyl (OH) group at the fourth carbon. - Reagent: Sodium borohydride (NaBH₄) in methanol. - Product: The ketone is reduced to a secondary alcohol (OH group on the second carbon), keeping the additional OH group unchanged. **Summary
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