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3 years ago

What major product is formed when heptyne is treated with Br2 in CCl4?

Chemistry

2 answers:

melisa1 [442]3 years ago

8 0

Answer:

Option A is the Answer. 

Explanation:

Treatment of alkyne with Bromine in the presence of $CCl_4$ produces a trans product that is, a alkenyl bromide.

$CH_3 \ CH_2 \ CH_2 \ CH_2 \ CH_2 - C = CH \frac {(CCl_4)}{(Br_2 )} > CH_3 \ CH_2 \ CH_2\ CH_2 \CH_2\ -CBr = CHBr$

E-1, 2-dibromoheptene

It is a trans intermediate on further halogenations produces a tetrabromopentane.

It is an electrophilic expansion response in which the triple bond breaks to turn into a twofold bond and creates a dibromoalkene (E-1,2 dibromoheptene), and on further expansion, the twofold bond turns into a solitary bond and delivers a tetrabromoalkane.

ANTONII [103]3 years ago

4 0

Answer:A 1,2-dibromoheptene

Explanation:

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2h3po4 + 3mg(oh)2 → mg3(po4)2 + 6h2o phosphoric acid, h3po4, is neutralized by magnesium hydroxide, mg(oh)2, according to the eq

Rina8888 [55]

The balanced equation for the neutralisation reaction is as follows
2H₃PO₄ + 3Mg(OH)₂ --> Mg₃(PO₄)₂ + 6H₂O
stoichiometry of H₃PO₄ to H₂O is 2:6
number of H₃PO₄ moles reacted - 0.24 mol
if 2 mol of H₃PO₄ form 6 mol of H₂O
then 0.24 mol of H₃PO₄ forms - 6/2 x 0.24 = 0.72 mol of H₂O
therefore 0.72 mol of H₂O are formed

4 0

4 years ago

Explain specifically how an electron gives off light in an atom.

exis [7]

Answer:

Then, at some point, these higher energy electrons give up their "extra" energy in the form of a photon of light, and fall back down to their original energy level.

Explanation:

When properly stimulated, electrons in these materials move from a lower level of energy up to a higher level of energy and occupy a different orbital.

3 0

3 years ago

John dissolves .5g of a white powder in 25g of benzene (FP 5oC) (kf benzene is 5.1) and finds the solution freezes at 3.7oC. Det

navik [9.2K]

Answer:

The compound has a molar mass of 78.4 g/mol

Explanation:

Step 1: data given

Mass of a sample = 0.5 grams

Mass of benzene = 25 grams

Freezing poing = 5 °C

Kf of benzene = 5.1 °C/m

Freezing point solution = 3.7 °C

Step 2: Calculate molality

ΔT = i*Kf*m

⇒with ΔT = the freezing point depression = 5.0 - 3.7 = 1.3 °C

⇒with i = the can't hoff factor = 1

⇒with Kf = the freezing point depression constant of benzene = 5.1 °C/m

⇒with m = the molality

1.3 = 5.1 * m

m = 1.3 / 5.1

m = 0.255 moles /kg

Step 3: Calculate moles

Molality = moles / mass benzene

0.255 molal = moles / 0.025 kg

Moles = 0.255 molal * 0.025 kg

Moles = 0.006375 moles

Step 4: Calculate molar mass of the compound

Molar mass compund = mass / moles

Molar mass compound = 0.5 grams / 0.006375 moles

Molar mass compound = 78.4 g/mol

The compound has a molar mass of 78.4 g/mol

7 0

3 years ago

Convert 10 liter into m3.

stellarik [79]

0.01 cubic meters
Hope this helps

3 0

3 years ago

Read 2 more answers

In which set do all elements tend to form anions in binary ionic compounds? li, na, k n, o, i c, s, pb k, fe, br?

pav-90 [236]

Atoms or molecule after gaining of electron possesses negative charge and is known as anion.

For the given sets:

$Li, Na, K$

The given elements are alkali metals and have tendency to lose electrons easily and form cations.

$N, O, I$

The given elements are non-metals and are electronegative. So, they gain electrons easily and form anion.

$C, S, Pb$

Carbon has tendency to form bond by sharing of electrons, Sulfur has tendency to gain electrons and form anion whereas Lead has tendency to lose electron.

$K, Fe, Br$

Potassium and Iron has tendency to lose electron and form cation whereas Bromine has tendency to gain electron to form anion.

Hence, from the given sets, all elements of set: $N, O, I$ have tendency to form anions in binary ionic compounds.

8 0

3 years ago

Read 2 more answers