The Nucleophilic substitution of Halo alkanes
The question is not very specific, so there is more than just one answer, but I'm assuming you are referring to a radical bromination of an alkane (ethane) versus an electrophilic bromination of an alkene (ethene).Br2 in the presence of a radical initiator (such as light or heat) will add to ethane to form 2-bromoethane as the major product in a radical mechanism. This goes through an initiation step (forming 2 bromine radicals), followed by propagation to the alkane (forming a secondary ethyl radical), followed by a termination step. The termination step leading to the product is one where another bromine radical joins with the ethyl radical.In the absence of light or heat, bromine cannot react with an alkane, but it can react as an electrophile with an alkene. In this type of reaction (electrophilic addition to an alkene), the ∏-bond (double bond) on ethene attacks a bromine atom (from Br2) and kicks out a bromide (Br-). The bromine that was just added forms two bonds (one on each carbon of the double bond), giving a three-membered C-Br-C ring called a bromonium ion (since the bromine atom now has a positive charge). The bromide that left before can now attack the backside of the bromonium ion, opening the 3-membered ring, and adding anti to form a dibromoalkane (1,2-dibromoethane in this example). This reaction is stereospecific because in the major product the bromine atoms will always add anti (to the opposite side) on the alkene.
When chlorine gas (Cl2) reacts with ethyne (C2H2), it undergoes an addition reaction, resulting in the formation of 1,2-dichloroethane (C2H2Cl2). The triple bond in ethyne is broken as chlorine atoms add across the carbon-carbon bond. This reaction typically occurs in the presence of light or heat, facilitating the addition process. Overall, the reaction transforms the alkyne into a more saturated compound.
Oxygen is much more dense than ethyne. This is why a test tube full of ethyne would be held upside down so the ethyne doesnt escape into the air.
The chemical formula of ethyne (acerylene) is C2H2 - two hydrogen atoms.
Ethyne
Yes, ethyne (acetylene) can react with bromine to form 1,2-dibromoethane. This is an addition reaction where the bromine atoms add across the carbon-carbon triple bond in ethyne.
One way to distinguish between ethene and ethyne is by performing a bromine water test. Ethene will decolorize bromine water, turning it from orange to colorless, while ethyne will not react with bromine water. This test takes advantage of the unsaturation in ethene that allows it to quickly react with bromine.
CH (triple bond) CH + Br2 -> BrC (triple bond) CBr
ethyne (or acetylene) is non polar
The balanced equation for the complete combustion of ethyne (C2H2) is: 2C2H2 + 5O2 -> 4CO2 + 2H2O
It is the balanced equation for the combustion of acetylene (or ethyne).
The question is not very specific, so there is more than just one answer, but I'm assuming you are referring to a radical bromination of an alkane (ethane) versus an electrophilic bromination of an alkene (ethene).Br2 in the presence of a radical initiator (such as light or heat) will add to ethane to form 2-bromoethane as the major product in a radical mechanism. This goes through an initiation step (forming 2 bromine radicals), followed by propagation to the alkane (forming a secondary ethyl radical), followed by a termination step. The termination step leading to the product is one where another bromine radical joins with the ethyl radical.In the absence of light or heat, bromine cannot react with an alkane, but it can react as an electrophile with an alkene. In this type of reaction (electrophilic addition to an alkene), the ∏-bond (double bond) on ethene attacks a bromine atom (from Br2) and kicks out a bromide (Br-). The bromine that was just added forms two bonds (one on each carbon of the double bond), giving a three-membered C-Br-C ring called a bromonium ion (since the bromine atom now has a positive charge). The bromide that left before can now attack the backside of the bromonium ion, opening the 3-membered ring, and adding anti to form a dibromoalkane (1,2-dibromoethane in this example). This reaction is stereospecific because in the major product the bromine atoms will always add anti (to the opposite side) on the alkene.
Benzene can be prepared from ethyne through a process called electrophilic aromatic substitution. In this process, ethyne reacts with a strong acid catalyst such as concentrated sulfuric acid to form benzene. The high temperature required for this reaction causes the triple bond in ethyne to break and the resulting carbons bond to form a benzene ring.
Chloroform will convert to ethylene by heating the former with silver powder. Chloroform is represented by the chemical formula CHCl3. Ethyne also known as acetylene is represented by the chemical formula C2H2. The silver powder acts as reducing agent in the given conversion, it reduces carbon and itself gets oxidized. The reaction for conversion is as follows: 2CHCl3 + 6Ag --> C2H2 + 6AgCl. In the given reaction, Ag is silver and AgCl is silver chloride.
To prepare 1-hexyne from ethyne, you would first need to convert ethyne to 1-bromohexane through a series of reactions. Ethyne would undergo bromination to form 1,2-dibromoethane, followed by a substitution reaction with sodium cyanide to form 1-cyano-1,2-dibromoethane. Finally, hydrolysis of the nitrile group would yield 1-hexyne. This multi-step process allows for the conversion of ethyne to 1-hexyne through intermediate compounds.
When ethyne is passed through ammoniacal cuprous chloride solution, the blue color of the cuprous chloride solution fades as the ethyne reduces cuprous chloride to copper, forming reddish-brown copper. This reaction is a test for unsaturation in organic compounds.
By combustion ethyne is transformed in carbon dioxide and water.