Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
Alright, let's break down the expected products for the [tex]\( \text{S}_\text{N}2 \)[/tex] reaction of 1-bromobutane with each of the following reagents:
### Part (a): [tex]\( \text{NaI} \)[/tex] (Sodium Iodide)
1-bromobutane reacts with sodium iodide in an [tex]\( \text{S}_\text{N}2 \)[/tex] reaction, where iodide ion ([tex]\( \text{I}^- \)[/tex]) acts as the nucleophile. In an [tex]\( \text{S}_\text{N}2 \)[/tex] reaction, the nucleophile attacks the carbon attached to the leaving group (bromine in this case) from the opposite side, displacing the bromide ion ([tex]\( \text{Br}^- \)[/tex]).
Product:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{I} \][/tex]
(butane to 1-iodobutane)
### Part (b): [tex]\( \text{KOH} \)[/tex] (Potassium Hydroxide)
When 1-bromobutane reacts with potassium hydroxide in an [tex]\( \text{S}_\text{N}2 \)[/tex] reaction, the hydroxide ion ([tex]\( \text{OH}^- \)[/tex]) acts as the nucleophile. The hydroxide attacks the carbon attached to the bromine, displacing the bromide ion.
Product:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \][/tex]
(butane to 1-butanol)
### Part (c): [tex]\( \text{NH}_3 \)[/tex] (Ammonia)
In this reaction, ammonia ([tex]\( \text{NH}_3 \)[/tex]) acts as the nucleophile. The nitrogen in ammonia attacks the carbon attached to the bromine in an [tex]\( \text{S}_\text{N}2 \)[/tex] reaction, displacing the bromide ion. The initial product is substituted butane attached with an ammonium group, which might further lose a proton to form a primary amine.
Product:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{NH}_3^+ \][/tex]
This often follows with:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{NH}_2 \][/tex]
(butane to 1-butylamine)
### Part (d): [tex]\( \text{H - C} \equiv \text{C} - \text{Li} \)[/tex] (Ethynyl Lithium)
In this case, the ethynyl anion [tex]\( \left(\text{HC} \equiv \text{C}^-\right) \)[/tex] is the nucleophile. It will attack the carbon attached to the bromine in an [tex]\( \text{S}_\text{N}2 \)[/tex] reaction, displacing the bromide ion.
Product:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{C} \equiv \text{CH} \][/tex]
(butane to 1-pentyne)
In summary, the expected products from the [tex]\( \text{S}_\text{N}2 \)[/tex] reactions of 1-bromobutane with the given reagents are:
(a) Sodium iodide [tex]\( \left(\text{NaI}\right) \)[/tex]: 1-iodobutane [tex]\( \left(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{I}\right) \)[/tex]
(b) Potassium hydroxide [tex]\( \left(\text{KOH}\right) \)[/tex]: 1-butanol [tex]\( \left(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\right) \)[/tex]
(c) Ammonia [tex]\( \left(\text{NH}_3\right) \)[/tex]: 1-butylamine [tex]\( \left(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{NH}_2\right) \)[/tex]
(d) Ethynyl lithium [tex]\( \left(\text{H - C} \equiv \text{C} - \text{Li}\right) \)[/tex]: 1-pentyne [tex]\( \left(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{C} \equiv \text{CH}\right) \)[/tex]
### Part (a): [tex]\( \text{NaI} \)[/tex] (Sodium Iodide)
1-bromobutane reacts with sodium iodide in an [tex]\( \text{S}_\text{N}2 \)[/tex] reaction, where iodide ion ([tex]\( \text{I}^- \)[/tex]) acts as the nucleophile. In an [tex]\( \text{S}_\text{N}2 \)[/tex] reaction, the nucleophile attacks the carbon attached to the leaving group (bromine in this case) from the opposite side, displacing the bromide ion ([tex]\( \text{Br}^- \)[/tex]).
Product:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{I} \][/tex]
(butane to 1-iodobutane)
### Part (b): [tex]\( \text{KOH} \)[/tex] (Potassium Hydroxide)
When 1-bromobutane reacts with potassium hydroxide in an [tex]\( \text{S}_\text{N}2 \)[/tex] reaction, the hydroxide ion ([tex]\( \text{OH}^- \)[/tex]) acts as the nucleophile. The hydroxide attacks the carbon attached to the bromine, displacing the bromide ion.
Product:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \][/tex]
(butane to 1-butanol)
### Part (c): [tex]\( \text{NH}_3 \)[/tex] (Ammonia)
In this reaction, ammonia ([tex]\( \text{NH}_3 \)[/tex]) acts as the nucleophile. The nitrogen in ammonia attacks the carbon attached to the bromine in an [tex]\( \text{S}_\text{N}2 \)[/tex] reaction, displacing the bromide ion. The initial product is substituted butane attached with an ammonium group, which might further lose a proton to form a primary amine.
Product:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{NH}_3^+ \][/tex]
This often follows with:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{NH}_2 \][/tex]
(butane to 1-butylamine)
### Part (d): [tex]\( \text{H - C} \equiv \text{C} - \text{Li} \)[/tex] (Ethynyl Lithium)
In this case, the ethynyl anion [tex]\( \left(\text{HC} \equiv \text{C}^-\right) \)[/tex] is the nucleophile. It will attack the carbon attached to the bromine in an [tex]\( \text{S}_\text{N}2 \)[/tex] reaction, displacing the bromide ion.
Product:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{C} \equiv \text{CH} \][/tex]
(butane to 1-pentyne)
In summary, the expected products from the [tex]\( \text{S}_\text{N}2 \)[/tex] reactions of 1-bromobutane with the given reagents are:
(a) Sodium iodide [tex]\( \left(\text{NaI}\right) \)[/tex]: 1-iodobutane [tex]\( \left(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{I}\right) \)[/tex]
(b) Potassium hydroxide [tex]\( \left(\text{KOH}\right) \)[/tex]: 1-butanol [tex]\( \left(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\right) \)[/tex]
(c) Ammonia [tex]\( \left(\text{NH}_3\right) \)[/tex]: 1-butylamine [tex]\( \left(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{NH}_2\right) \)[/tex]
(d) Ethynyl lithium [tex]\( \left(\text{H - C} \equiv \text{C} - \text{Li}\right) \)[/tex]: 1-pentyne [tex]\( \left(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{C} \equiv \text{CH}\right) \)[/tex]
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.