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arlik [135]
3 years ago
5

For alkyl halides used in SN1 and SN2 mechanisms, rank the leaving groups in order of reaction rate. You are currently in a rank

ing module. Turn off browse mode or quick nav, Tab to move, Space or Enter to pick up, Tab to move items between bins, Arrow Keys to change the order of items, Space or Enter to drop.
Chemistry
1 answer:
Alex777 [14]3 years ago
4 0

Answer:

Iodide> Bromide > chloride > flouride

Explanation:

During a nucleophilic substitution reaction, a nucleophilie replaces another in a molecule.

This process may occur via an ionic mechanism (SN1) or via a concerted mechanism (SN2).

In either case, the ease of departure of the leaving group is determined by the nature of the C-X bond. The stronger the C-X bond, the worse the leaving group will be in nucleophilic substitution. The order of strength of C-X bond is F>Cl>Br>I.

Hence, iodine displays the weakest C-X bond strength and it is thus, a very good leaving group in nucleophillic substitution while fluorine displays a very high C-X bond strength hence it is a bad leaving group in nucleophilic substitution.

Therefore, the ease of the use of halide ions as leaving groups follows the trend; Iodide> Bromide > chloride > flouride

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Answer:

\boxed{\text{C. the equilibrium between the solid and its ions in solution}}

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Answer:

C ) 2, 1, 2

Explanation:

The given reaction is synthesis reaction in which lithium and bromine react to form lithium bromide.

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Li + Br₂    →    LiBr

Balanced chemical equation:

2Li + Br₂    →    2LiBr

Step 1:

Li + Br₂    →    LiBr

left hand side                         Right hand side

Li = 1                                        Li = 1

Br = 2                                        Br = 1

Step 2:

Li + Br₂    →    2LiBr

left hand side                         Right hand side

Li = 1                                        Li = 2

Br = 2                                        Br = 2

Step 3:

2Li + Br₂    →    2LiBr

left hand side                         Right hand side

Li = 2                                       Li = 2

Br = 2                                        Br = 2

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