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allsm [11]
4 years ago
11

Write a structural formula for the principal organic product formed in the reaction of methyl bromide with NaOH (sodium hydroxid

e). Click the "draw structure" button to launch the drawing utility.Write a structural formula for the principal organic product formed in the reaction of methyl bromide with NaOH (sodium hydroxide). Click the "draw structure" button to launch the drawing utility.

Chemistry
1 answer:
Naily [24]4 years ago
4 0

Answer:

Methanol is produced as product of the reaction.

Explanation:

Methyl bromide contains a good leaving group Br^{-} because of high polarizability of Br atom and weak C-Br bond.

OH^{-} is a strong base as well as strong nucleophile due to small size and high electronegativity of oxygen atom.

Here methyl bromide primarily acts as an electrophile instead of an acid because of it's very weak acidity.

So, OH^{-} acts as nucleophile towards methyl bromide by replacing bromide ion and forms methanol.

Reaction goes through S_{N}2 mechanism.

Structure formula of product has been given below.

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Elanso [62]

Answer: The pH of 0.10 M Cu(NO_{3})_{2}(aq) is 4.49.

Explanation:

Given: Initial concentration of Cu(H_{2}O)^{2+}_{6} = 0.10 M

K_{a} = 1.0 \times 10^{-8}

Let us assume that amount of Cu(H_{2}O)^{2+}_{6} dissociates is x. So, ICE table for dissociation of  Cu(H_{2}O)^{2+}_{6}  is as follows.

                               Cu(H_{2}O)^{2+}_{6} \rightleftharpoons [Cu(H_{2}O)_{5}(OH)]^{+} + H_{3}O^{+}

Initial:                       0.10 M                       0                       0

Change:                    -x                              +x                      +x

Equilibrium:           (0.10 - x) M                    x                        x

As the value of K_{a} is very small. So, it is assumed that the compound will dissociate very less. Hence, x << 0.10 M.

And, (0.10 - x) will be approximately equal to 0.10 M.

The expression for K_{a} value is as follows.

K_{a} = \frac{[Cu(H_{2}O)^{2+}_{6}][H_{3}O^{+}]}{[Cu(H_{2}O)^{2+}_{6}]}\\1.0 \times 10^{-8} = \frac{x \times x}{0.10}\\x = 3.2 \times 10^{-5}

Hence, [H_{3}O^{+}] = 3.2 \times 10^{-5}

Formula to calculate pH is as follows.

pH = -log [H^{+}]

Substitute the values into above formula as follows.

pH = -log [H^{+}]\\= - log (3.2 \times 10^{-5})\\= 4.49

Thus, we can conclude that the pH of 0.10 M Cu(NO_{3})_{2}(aq) is 4.49.

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Explanation:

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