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Hatshy [7]
2 years ago
7

Compounds X and Y are both C6H13Cl compounds formed in the radical chlorination of 3-methylpentane. Both X and Y undergo base-pr

omoted E2 elimination to give a mixture of alkenes.In water X and Y each react to form a mixture of substitution and elimination products; however, X reacts much faster than does Y.What is the structure of Y?

Chemistry
1 answer:
zubka84 [21]2 years ago
6 0

Answer:

Y is a 3-chloro-3-methylpentane.

The structure is shown in the figure attached.

Explanation:

The radical chlorination of 3-methylpentane can lead to a tertiary substituted carbon (Y) and to a secondary one (X).

The E2 elimination mechanism, as shown in the figure, will happen with a simulyaneous attack from the base and elimination of the chlorine. This means that primary and secondary substracts undergo the E2 mechanism faster than tertiary substracts.

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Enough of a monoprotic weak acid is dissolved in water to produce a 0.0144 0.0144 M solution. The pH of the resulting solution i
alexdok [17]

Answer:

The value of dissociation constant of the monoprotic acid is 1.099\times 10^{-3}.

Explanation:

The pH of the solution = 2.46

pH=-\log[H^+]

2.46=-\log[H^+]

[H^+]=0.003467 M

HA\rightleftharpoons H^++A^-

Initially

0.0144         0      0

At equilibrium

(0.0144-x)       x       x

The expression if an dissociation constant is given by :

K_a=\frac{[A^-][H^+]}{[HA]}

K_a=\frac{x\times x}{(0.0144-x)}

x=[H^+]=0.003467 M

K_a=\frac{0.003467 \times 0.003467 }{(0.0144-0.003467 )}

K_a=1.099\times 10^{-3}

The value of dissociation constant of the monoprotic acid is 1.099\times 10^{-3}.

3 0
3 years ago
PLS HELP 6TH GRADE ITS ONE QUESTION ABOUT SCIENCE
julia-pushkina [17]

Answer: The answer is C

Explanation:

5 0
3 years ago
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73g of HCL in 2.00l of HCL solution
Whitepunk [10]
What is your question?
8 0
3 years ago
Calculate the % composition of the unknown liquid using your most precise result. It is a mixture of ethanol (D[ETOH] = 0.7890 g
Rufina [12.5K]

% composition of ethanol = 34.51%

% composition of water  = 65.49%

<h3>What is density?</h3>

A material's density is defined as its mass per unit volume.

Given data:

The density of ethanol = 0.7890 g/mL

The density of water = 0.9982 g/mL

The density of mixture = 0.926 g/mL

Let the % composition of ethanol = x

Let the % composition of water = 100-x

Now density of the mixture

\frac{Mass}{Volume}

Mass = \frac{percent  \;of  \;ethanol  \;X  \;density  \;of  \;ethanol  \;+  \\ \;percent  \;of  \;water X  \;density  \;of  \;water}{100}

0.926 = \frac{x X  0.7890 g/mL  \;+  (100-x) X  0.9982 g/mL}{100}

x= 34.51 %

Hence,

% composition of ethanol = 34.51%

% composition of water = 65.49%

Learn more about the density here:

brainly.com/question/952755

#SPJ1

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1 year ago
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cricket20 [7]
1- metal and non metal
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