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dybincka [34]
3 years ago
13

A certain alkyl halide is reacted with OH- to form an alcohol. The alkyl halide is optically active but the product(s) is/are op

tically inactive. Which of the following could be the reactant?a) 3-bromo-3-methylhexane.b) 1-chlorobutane.c) 2-bromo-2-methylbutane.d) 3-bromo-2,3,4-trimethypentane.
Chemistry
1 answer:
Snowcat [4.5K]3 years ago
8 0

Answer:

a. 3-brumo - 3-methylhexane

Explanation:

Alkyl Halides can undergo substitution reactions. Nucleophiles are electron rich species and has negative charge while Electrophiles are electron deficient species which carry positive charge. Alkyl halide which have polar carbon atom are electrophiles.

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Hey there!

Here is the answer to your question.

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I hope this helps you! Let me know if you need anymore help on anything you are struggling with! ☺

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The yellow precipitate formed in the reaction between KI and Pb(NO3)2 is . b. The white precipitate formed in the reaction betwe
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Answer:

a. Here, the yellow precipitate is lead (II) iodide:

2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)

b. Here the white precipitate is barium sulfate:

BaCl(aq)+H_2SO_4(aq)\rightarrow BaSO_3(s)+2HCl(aq)

c. Here the brown precipitate is iron (III) hydroxide:

3NaOH(aq)+FeCl_3(aq)\rightarrow 3NaCl(aq)+Fe(OH)_3(s)

d. Here, the precipitate is copper (II) hydroxide:

CuSO_4(aq)+2NaOH(aq)\rightarrow Cu(OH)_2(s)+Na_2SO_4(aq)

Explanation:

Hello.

In this case, since the precipitation reactions are characterized by the formation of a product which is highly insoluble in water, specially very heavy products having heavy metals at the cations, for each case we proceed as follows:

a. Here, the yellow precipitate is lead (II) iodide:

2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)

b. Here the white precipitate is barium sulfate:

BaCl(aq)+H_2SO_4(aq)\rightarrow BaSO_3(s)+2HCl(aq)

c. Here the brown precipitate is iron (III) hydroxide:

3NaOH(aq)+FeCl_3(aq)\rightarrow 3NaCl(aq)+Fe(OH)_3(s)

d. Here, the precipitate is copper (II) hydroxide:

CuSO_4(aq)+2NaOH(aq)\rightarrow Cu(OH)_2(s)+Na_2SO_4(aq)

Best regards!

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3 years ago
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