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3 years ago

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Compound A has molecular formula C5H11Br. When compound A is treated with bromine in the presence of UV light, the major product

is 2,2-dibromopentane. Treatment of compound A with NaSH (a strong nucleophile) produces a compound with one chirality center having the R configuration. What is the structure of compound A

1 answer:

damaskus [11]3 years ago

6 0

The compound (A) is 2-bromo-pentane. See attached picture for the chemical diagram.

Explanation:

2-bromo-pentane (A) reacts with bromine (Br₂) under UV light to form 2,2-dibromopentane and HBr. Also 2-bromo-pentane (A) reacts with NaSH to from pentane-2-thiol, which have an asimetric carbon, and NaBr.

You may find the chemical reactions and diagrams of the compounds in the attached picture.

Learn more about:

structure of organic compounds

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Glyceraldehyde-3-phosphate is an important intermediate molecule in the cell's metabolic pathways because __________.

olga55 [171]

Answer:

Present in both catabolic and anabolic pathways

Explanation:

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In glycolysis, Glyceraldehyde-3-phosphate is produced by breakdown of fructose-1,6 -bisphosphate. Further Glyceraldehyde-3-phosphate converted to pyruvate and pyruvate is further used in citric acid cycle for energy production. Therefore, it is used in catabolic pathway too.

Glyceraldehyde-3-phosphate is an important intermediate molecule in the cell's metabolic pathways because it is present in both catabolic and anabolic pathways.

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3 years ago

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!: (a) 20 cm: 80

RoseWind [281]

Answer:

THE LENGTH OF THE AIR COLUMN IS 9.5 CM

Explanation:

Taking the atmospheric pressure to be 760 mmHg;

When the capillary tube is held horizontally, the pressure of the tube is 760 mmHg

when the capillary tube is held vertically, the pressure increases by 4 cm = 40 mm

The new pressure of the tube is hence, 760 + 40 mmHg = 800 mmHg

Using the pressure forlmula;

P1 V1 = P2 V2

P1 A1 L1 = P2 A2 L2

where A1 and A2 is the area of the capillary tube and it is equal, it cancels out.

P1 l1 = P2 l2

l2 = P1 l1 / P2

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The length of the air in the tube is 9.5 cm.

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3 years ago

What is orbital theory of bonding in metals

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3 years ago

The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C

avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

$q=c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

c = specific heat = $5.20kJ/^oC$

$T_{final}$ = final temperature = $27.43^oC$

$T_{initial}$ = initial temperature = $22.93^oC$

Now put all the given values in the above formula, we get:

$q=5.20kJ/^oC\times (27.43-22.93)^oC$

$q=23.4kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = $\frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole$

$\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole$

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

$\Delta H=\Delta U+\Delta n_gRT$

or,

$\Delta U=\Delta H-\Delta n_gRT$

where,

$\Delta H$ = change in enthalpy = $-2805.8kJ/mol$

$\Delta U$ = change in internal energy = ?

$\Delta n_g$ = change in moles = 0 (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = $27.43^oC=273+27.43=300.43K$

Now put all the given values in the above formula, we get:

$\Delta U=\Delta H-\Delta n_gRT$

$\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K$

$\Delta U=-2805.8kJ/mol-0$

$\Delta U=-2805.8kJ/mol$

Therefore, the internal energy change is -2805.8 kJ/mol

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3 years ago