Ask question

Ask question

All categories

2 years ago

11

1. why is HCl not a good choice as the acid to catalyze a dehydration reaction?

1 answer:

Len [333]2 years ago

7 0

Answer:

1) HCl contains the Cl^- which is a good nucleophile

2) 2-methyl-2- heptanol > 2-heptanol > 1-heptanol

3) see image attached

Explanation:

If the dehydration of alcohols is carried out using HCl, the chloride ion which is a good nucleophile will attack the substrate to yield an undesirable product.

The dehydration of alcohols is an E1 reaction. Recall that the ease of E1 reaction increases in the order 3°> 2°> 1°. Hence, 2-methyl-2- heptanol forms a tertiary carbocation intermediate during dehydration and has the greatest ease of dehydration.

The three products formed during the dehydration of 3,3-dimethyl-2-butanol are shown in the image attached. Two out of the three are formed by rearrangement reactions.

You might be interested in

A sample of gas initially has a volume of 1.42 L at 293 K and 1.30 atm. What volume will the sample have if the pressure changes

worty [1.4K]

Answer:

1.35

Explanation:

5 0

2 years ago

What is N? (Needed electrons)

lianna [129]

Do you know what is meant by needed? does it mean lone pairs?

8 0

3 years ago

When Iron (III) Choride is added to a solution of synthesized aspirin what is the method of detection

zubka84 [21]

Answer:

seeing the color change

Explanation:

3 0

2 years ago

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

<u>Answer:</u> The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

4 0

3 years ago

Please help fast, I will give brainliest.

Serga [27]

Answer:

B

Explanation:

7 0

2 years ago

Read 2 more answers