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3 years ago

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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Many repeating, simple subunits are joined together. What is the most likely

Hatshy [7]

Answer:

O a polymer

Explanation:

When many repeating simple subunits are joined together, this results into a polymer.

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3 years ago

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Lapatulllka [165]

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2 years ago

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stiv31 [10]

Answer:

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3 years ago

The expected o-c-h angle in this molecule is degrees. the expected hybridization at the central carbon is

WINSTONCH [101]

Each neutral carbon atom contains four valence electrons and may form up to four electron domains. Possible hybridizations include

$sp^{3}$ , four electron domains, as in ethane $\text{C}_2\text{H}_6$
$sp^{2}$ , three electron domains, as in ethene $\text{C}_2\text{H}_4$
$sp$ , two electron domains, as in ethyne $\text{C}_2\text{H}_2$

Molecules of each of the three hybridization demonstrate spatial configurations that would maximizes the separation between the electron domains.

Carbon atoms with a $sp^{3}$ hybridization would demonstrate a tetrahedral configuration with a bond angle of approximately $109.5\textdegree{}$
Carbon atoms with a $sp^{2}$ hybridization would demonstrate a triangular planar configuration with a bond angle of $120\textdegree{}$
Carbon atoms with a $sp$ hybridization would demonstrate a linear configuration with a bond angle of $180\textdegree{}$

Bond angles are characteristic of the spatial configuration of electron domains and identifies the hybridization of the central carbon atom.

Note that each hydrogen atom contains only one valence electron and would form only single bonds. It takes two valence electrons for oxygen atoms to achieve an octet such that each oxygen form only two bonds at a single time. Therefore given the fact that the carbon is bonded to both hydrogen and oxygen, only the following hybridizations are possible

$sp^{3}$ in which the oxygen atom forms a carbon-oxygen double bond with the central carbon atom;
$sp^{2}$ in which the oxygen atom forms a single bond with the central carbon atom and with a second atom.

3 0

3 years ago

When 50.0 mL of 1.27 M of HCl(aq) is combined with 50.0 mL of 1.32 M of NaOH(aq) in a coffee-cup calorimeter, the temperature of

sergij07 [2.7K]

Answer:

-55.9kJ/mol is the change in enthalpy of the reaction

Explanation:

In the reaction:

HCl(aq) + NaOH(aq) → H₂O(l) + NaCl

Some heat is released per mole of reaction.

To know how many moles reacts we need to find limiting reactant:

Moles HCl = 0.050L ₓ (1.27mol / L) = 0.0635 moles HCl

Moles NaOH = 0.050L ₓ (1.32mol / L) = 0.066 moles NaOH

As there are more moles of NaOH than moles of HCl, HCl is limiting reactant and moles of reaction are moles of limiting reactant, 0.0635 moles

Using the coffee-cup calorimeter equation we can find how many heat was released thus:

Q = C×m×ΔT

Where Q is heat released, C is specific heat of the solution (4.18J/g°C), m is mass of solution (100g because there are 100mL of solution -50.0mL of HCl and 50.0mL of NaOH- and density is 1g/mL) and ΔT is change in temperature (8.49°C)

Replacing:

Q = 4.18J/g°C×100g×8.49°C

Q = 3548.8J of heat are released in the reaction

Now, change in enthalpy, ΔH, is equal to change in heat (As is released heat ΔH < 0) per mole of reaction, that is:

ΔH = Heat / mol of reaction

ΔH = -3548.8J / 0.0635 moles of reaction

Negative because is released heat.

ΔH = -55887J / mol

ΔH =

<h3>-55.9kJ/mol is the change in enthalpy of the reaction</h3>

3 0

2 years ago