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m_a_m_a [10]
3 years ago
10

What is the major organic product obtained from the following sequence of reactions? PhCH2CHO PhCH2CH2CHO PhCH2CH2COOH PhCH2COOH

Chemistry
1 answer:
yarga [219]3 years ago
8 0

Answer:

PhCH2CH2COOH

Explanation:

This is a reaction of PhCH2CH2Br with KCN in the presence of H3O^+. The reaction first leads to the formation of PhCH2CH2CN.

We must recall that part of the properties of nitriles is that they can be converted to carboxylic acids in the presence of H3O^+. This is a common synthetic route for carboxylic acids.

Therefore, when the PhCH2CH2CN is now further reacted with H3O^+, the carboxylic acid PhCH2CH2COOH is formed as the major organic product of the reaction, hence the answer given above.

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Answer:

It's explained below.

Explanation:

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3 years ago
A mixture containing 20 mole % butane, 35 mole % pentane and rest
notka56 [123]

Answer:

2.5 % butane, 42.2 % pentane and 55.3 % hexane

Explanation:

Hello,

In this case, the mass balance for each substance is given by:

Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB

Whereas y accounts for the fractions at the outlet distillate and x for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

0.9=\frac{y_bD}{z_bF}

So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:

y_bD=0.9*z_bF=0.9*0.2*100mol=18mol

The total distillate flow:

y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol

And the total bottoms flow:

F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025

Then, the molar fraction of pentane and hexane:

x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422

x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553

Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.

NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.

Regards.

8 0
3 years ago
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4 0
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